Does Normal Force change with new Force diagram?

[PacFan01]

All degrees are measured from the positive x-axis.

F^G = Force of gravity
F^N = Normal force
F^T = Force of tension

Problem:

A skier is being pulled up a 15° incline by a rope. I'm solving for F^T. The only forces acting on the skier are F^G, F^N, and F^T.

Attempt at a solution:

Initially, my free-body diagram shows the F^G at 270°, F^N at 105°, and F^T at 15°.

Let's say I redraw my free-body diagram such that F^G points at 255° F^N at 90°, and F^T at 0° (or 360°).

F_y^G =-mgcosθ. What I want to know is, does my F_y^N = +mgcosθ, or just +mg?

 

[Matterwave]

If the angle is 90 degrees, what's cosine of 90 degrees?

Make sure you don't just blindly apply equations, you must know where those equations come from. If you rotate your diagram, changing the "theta", your equations must necessarily take this into account.

 

[PacFan01]

Thanks for the reply, Matterwave.

I understand what you are saying. It's hard for me to look at a straight line (like in the case of F^G at 270°) and apply trigonometric values.

I can see that if we correlate sinθ with the y-axis, and cosθ with the x-axis, that F^G at 270° = mgsinθ where sin(270°) = -1, thus we typically refer to F^G as -mg, associating the negative sign with the direction of gravitational acceleration.

Thus, the answer to my question is F_y^N = +mgcosθ.

 

[CWatters]

I understand what you are saying. It's hard for me to look at a straight line (like in the case of FG at 270°) and apply trigonometric values.

Think to yourself... What is the component of F^G that acts in the F^T direction? and what is the component of F^G that acts at 90 degrees to F^T. Draw these it on your diagram and you will see they form a right angle triangle. Only then apply the trig.

Personally I don't think redrawing the diagram rotated 15 degrees makes it easier to understand.

 

[HalfLostAndSearching]

In general, the normal force (FN) is the force that a surface exerts on an object in contact with it, perpendicular to the surface. In this scenario, the normal force is the force that the incline exerts on the skier, preventing them from falling through the surface.

In the first free-body diagram, the normal force (FN) is represented as +mgcosθ, where θ is the angle of the incline. This is because the force of gravity (FG) is acting at an angle of 270°, and the normal force is perpendicular to the surface at an angle of 105°. So, the normal force is equal to the component of the force of gravity in the direction perpendicular to the surface, which is mgcosθ.

In the second free-body diagram, the angle of the incline has changed, but the force of gravity (FG) is still acting at an angle of 270°. Therefore, the normal force (FN) will also change in direction, but it will still be equal to the component of the force of gravity in the direction perpendicular to the surface, which is still mgcosθ. So, in this case, FyN = +mgcosθ, not just +mg.

In general, the normal force will change with a new force diagram if the angle of the incline changes, as the component of the force of gravity in the direction perpendicular to the surface will also change. However, if the angle of the incline remains the same, the normal force will also remain the same.